Concordia MATH 206 tutoring

Factor equation A

Equation A factors as $(x-5)(x+1)=0$, so its solutions are $5$ and $-1$.

Keep the original logarithm domain

For B, first require both logarithm arguments to be positive: $x-1>0$ and $x-3>0$. Together these conditions give $x>3$.

Combine the logarithms on that domain

On that domain, the product rule for logarithms gives$$\log_2((x-1)(x-3))=3.$$Therefore $(x-1)(x-3)=2^3=8$, which simplifies to equation A.

The candidates are again $5$ and $-1$, but only $5$ satisfies $x>3$. Check it in B: $\log_2 4+\log_2 2=2+1=3$.

Check candidates in the original equation

At $x=-1$, the original logarithms have negative arguments. A positive product does not make those individual logarithms defined. Preserve the original domain when combining them.